Question

Sum of three consecutive terms of an ap is 21 and sum of their squares is 165 find these terms

Answer 1

let the terms be a-d, a, a+d

sum=21

$a - d + a + a + d = 21 \\ 3a = 21 \\ a = 7...(1) \\ \\ sum \: of \: squares \: = 165 \\ (a - d) {}^{2} + {a}^{2} + (a + d) {}^{2} = 165\\ 3a {}^{2} + 2 {d}^{2} = 165 \\ 3(7 {}^{2} ) - 165 = - 2 {d}^{2} \\ d {}^{2} = 9 \\ d = 3 \: or \: - 3$



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Answer 2

Here is your solution

Let,
The terms be

a-d, a, a+d 

sum = 21

a - d + a + a + d = 21
3a = 21
a = 7...(1)

sum of squares = 165
(a - d)^2+ a^2 + (a + d)^2 = 165
3a^2+ 2d^2 = 165
3(7^2) - 165 = - 2d^2
d^2= 9
d = 3 , - 3